Birkhoff prime ideal theorem

Birkhoff Prime Ideal Theorem. Let L be a distributive latticeMathworldPlanetmath and I a proper lattice ideal of L. Pick any element aI. Then there is a prime idealMathworldPlanetmathPlanetmath P in L such that IP and aP.


If I is prime, then we are done. Let S:={JJ is an ideal in L, and aJ}. Then IS. Order S by inclusion. This turns S into a poset. Let C be a chain in S. Let K=C. If x,yK, then xJ1 and yJ2 for some ideals J1,J2C. Since C is a chain, we may assume that J1J2, so that xJ2 as well. This means xyJ2K. Next, assume xK and yx. Then xJ for some ideal JC, so that yJK also. This shows that K is an ideal. If aK, then aJ for some JCS, contradicting the definition of S. So aK and KS also. This shows that every chain in S has an upper bound. We can now appeal to Zorn’s lemma, and conclude that S has a maximal elementMathworldPlanetmath, say P.

We now want to show that P is the candidate that we are seeking: P is a prime ideal in L and aP. Since PS, P is an ideal such that aP. So the only thing left to prove is that P is prime. This amounts to showing that if xyP, then xP or yP. Suppose not: x,yP. Let Q1 be the ideal generated by elements of P and x, and Q2 the ideal generated by P and y. Since Q1 and Q2 properly contain P, aQ1 and aQ2. Write ap1x and ap2y, where p1,p2P. Then ap2(p1p2)x and ap1(p1p2)y. Take the meet of these two expressions, and we obtain (ap2)(ap1)((p1p2)x)((p1p2)y). Since L is distributive, on the left hand side, we get a(p1p2). On the right hand side, we have (p1p2)(xy)P. As the left hand side is less than or equal to the right hand side, we get that a(p1p2)P. Since aa(p1p2)P, aP, a contradictionMathworldPlanetmathPlanetmath. Therefore, P is prime and the proof is completePlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath. ∎

In the proof, we use the fact that, an element aL belongs to the ideal generated by ideals Ik iff a is less than or equal to a finite join of elements, each of which belongs to some Ik.


  1. 1.

    The theoremMathworldPlanetmath can be generalized: if we use a subset SI= instead of an element aI, there is a prime ideal P containing I but excluding S.

  2. 2.

    Birkhoff’s prime ideal theorem has been shown to be equivalentMathworldPlanetmathPlanetmathPlanetmathPlanetmath to the axiom of choiceMathworldPlanetmath, under ZF.

Title Birkhoff prime ideal theorem
Canonical name BirkhoffPrimeIdealTheorem
Date of creation 2013-03-22 17:02:18
Last modified on 2013-03-22 17:02:18
Owner CWoo (3771)
Last modified by CWoo (3771)
Numerical id 11
Author CWoo (3771)
Entry type Theorem
Classification msc 06D05
Classification msc 03E25
Related topic DistributiveLattice